Example#2: Baking Profits. Sara Lee Corp., maker of food, beverage, and household products, isknown especially for its baked products, marketed under its corporate name. For the five years endingJuly 1 of each year from 2002 to 2006, their bakery division reported the following profits.Time Year

Profits($M) MA(3)=

2002

140

140

2003

120

134

140

2004

156

138.67

137.3

134

2005

160

145.33

138.67

144.11

137.3

2006

150

155.33

145.33

145.88

144.11

2007

155

155

155.33

148.614

145.88

2008

130

145

155

143.03

148.614

2009

150

145

145

145.121

143.03

2010

Forecast using MA(3)

F t=

145

SES(0.3)

Forecast usingSES(0.3)

145.121

1) Plot the data. Which time series components seems to be present?(attach JMP graph)

The graph appears to have only an irregular

component

Stat252 fall 2016

Lab#9 take-home Ch:19.3-19.42) Find a 3 point moving average (. Enter in the column MA(3).

3) Use a 3-year moving average to predict profits for 2010. Because 2010 is current the mostaccurate forecast we can make is the prediction for 2009 of $145M4) Use a 8 year moving average to predict profits for 2010. Because we dont have actualprofits for 2010 we cannot perform an 8-point moving average5) Find a single exponential smoothing (with alpha=0.3. Enter in the column SES(0.3). See abovecolumn6) Predict the profit for 2010 using a single exponential smoothing with alpha=0.3 The mostaccurate (and still not very reliable) prediction we can make is the previous years forecastof $145.121M7) If the actual profits for 2010 were 140. Report the forecast errors for year 2010 using MA(3) andSES(0.3) methods. Just based on these forecast errors, which model gives a smaller forecasterror for 2010. Assuming actual value of 140: MA(3) = 140 SES(0.3) = 143.585 Error(MA(3)) =-5 Error(SES(0.3)) = -5.121. The MA(3) model provides a smaller forecast error8) Which method (MA(3) or SES(0.3)) provides a more accurate prediction for the last three years(m=3). MAD(MA(3)) = (25 + 5 + 5)/3 = 11.67 MSE(MA(3)) = (25^2 + 5^2 + 5^2)/3 = 225MAPE(MA(3)) =(100)(1/3)((25/130) + (5/150) + (5/140)) = 8.712 MAD(SES(0.3)) = (18.614 +6.97 + 5.121)/3 = 10.235 MSE(SES(0.3)) = (18.614^2 + 6.97^2 + 5.121^2)/3 = 140.429MAPE(SES(0.3)) =(100)(1/3)((18.614/130) + (6.97/150) + (5.121/140)) = 7.541 The SES(0.3)model has more accurate predictions as its mean errors are all lower than the MA(3) model.

Stat252 fall 2016

Lab#9 take-home Ch:19.3-19.4

c) Use JMP to compute the Durbin-Watson statistic and comment. (Just want you to practice the Durbin-Watsontest and dont worry about all other assumptions).

D = 0.813 k = 1 n = 17 dL = 0.873 dU = 1.102

Based on these numbers we would have evidence of a positiveautocorrelation, meaning that the errors are not independentd) Based on your answer in part c), what do you conclude from this test about Model #1.We would reject the null hypothesis and conclude that errors are not independente) Create a Model#2, accounting for the seasonal component. Attach the output.

Model 2On the surface Model 2 appears to be the better choice because ithas a higher R^2adj of 94.5% and a lower Se of 2.365.

f) Interpret the coefficient for the variable Time.

For each additional day the average earnings for Harry Potter will decrease by $1.429(Billions? Millions?)g) Interpret the coefficient for the variable Saturday.b(Saturday) = 4.571 The average increase in earnings for Harry Potter is $4.571(Billions? Millions?) forSaturdays when compared to Sundays, after accounting for time.